Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ is a $p-$group and $H$ a proper subgroup of $G$, $|H|=p^s$. Show that $G$ has a subgroup $K$ such that $H \subset K$ and $|K|=p^{s+1}$.

I try to find a subgroup $K$ of $G$, $|K|=p^t$ with some $t>s$. Let $K=\langle g,H\rangle$ with some $g\in G,g\not\in H$, we have ${|g|=p^n,|H|=p^s}\Rightarrow|K|=p^n\times p^s$. Is this true? Can you help me!

share|cite|improve this question

For this we can use that a $p$-group has a non-trivial center and proceed by induction on the order. If the subgroup does not contain the center then its normalizer does and the normalizer is therefore a strictly larger subgroup. If it does contain the center then you can mod out by a central subgroup of order $p$ and use the induction.

share|cite|improve this answer
@ Tobias: Ok, a finite $p-$group $G$ has a non-trivial center. But is it true in case G is infinite? – tlquyen Jan 2 '13 at 14:28
The result you want to show does not hold for arbitrary $p$-groups. For example, there exists an infinite group where all proper subgroups have order $p$ for a fixed prime $p$. – Tobias Kildetoft Jan 2 '13 at 15:16
Oh, this is an exercise in my final test. Can you give me some details about the example that you say about. Thank you! – tlquyen Jan 2 '13 at 15:44
The infinite groups I mentioned are called the Tarski monster groups (there is a short article about them on Wikipedia). – Tobias Kildetoft Jan 2 '13 at 15:47
Tobias, thank you very much. – tlquyen Jan 2 '13 at 16:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.